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Giuga Numbers

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Given a number N, the task is to check if N is an Giuga Number or not. If N is an Giuga Number then print “Yes” else print “No”.

A Giuga Number is a composite number N such that p divides (N/p – 1) for every prime factor p of N
 

Examples:  

Input: N = 30 
Output: Yes 
Explanation: 
30 is a composite number whose prime divisors are {2, 3, 5}, such that 
2 divides 30/2 – 1 = 14, 
3 divides 30/3 – 1 = 9, and 
5 divides 30/5 – 1 = 5.

Input: N = 161 
Output: No 

Approach: The idea is to check if N is composite number or not. If not then print “No”. 
If N is a composite number then find the prime factors of a number and for each prime factor p check if the condition p divides (n/p – 1) holds true or not. If the above condition holds true then print “Yes” else print “No”.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if n is
// a composite number
bool isComposite(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return false;
 
    // This is checked to skip
    // middle 5 numbers
    if (n % 2 == 0 || n % 3 == 0)
        return true;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return true;
 
    return false;
}
 
// Function to check if N is a
// Giuga Number
bool isGiugaNum(int n)
{
    // N should be composite to be
    // a Giuga Number
    if (!(isComposite(n)))
        return false;
 
    int N = n;
 
    // Print the number of 2s
    // that divide n
    while (n % 2 == 0) {
 
        if ((N / 2 - 1) % 2 != 0)
            return false;
 
        n = n / 2;
    }
 
    // N must be odd at this point.
    // So we can skip one element
    for (int i = 3; i <= sqrt(n);
         i = i + 2) {
 
        // While i divides n,
        // print i and divide n
        while (n % i == 0) {
            if ((N / i - 1) % i != 0)
                return false;
 
            n = n / i;
        }
    }
 
    // This condition is to handle
    // the case when n
    // is a prime number > 2
    if (n > 2)
        if ((N / n - 1) % n != 0)
            return false;
 
    return true;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 30;
 
    // Function Call
    if (isGiugaNum(N))
        cout << "Yes";
    else
        cout << "No";
}


Java




// Java program for the above approach
class GFG{
 
// Function to check if n is
// a composite number
static boolean isComposite(int n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return false;
 
    // This is checked to skip
    // middle 5 numbers
    if (n % 2 == 0 || n % 3 == 0)
        return true;
 
    for(int i = 5; i * i <= n; i = i + 6)
       if (n % i == 0 || n % (i + 2) == 0)
           return true;
 
    return false;
}
 
// Function to check if N is a
// Giuga Number
static boolean isGiugaNum(int n)
{
     
    // N should be composite to be
    // a Giuga Number
    if (!(isComposite(n)))
        return false;
 
    int N = n;
 
    // Print the number of 2s
    // that divide n
    while (n % 2 == 0)
    {
        if ((N / 2 - 1) % 2 != 0)
            return false;
        n = n / 2;
    }
 
    // N must be odd at this point.
    // So we can skip one element
    for(int i = 3; i <= Math.sqrt(n);
                   i = i + 2)
    {
        
       // While i divides n,
       // print i and divide n
       while (n % i == 0)
       {
           if ((N / i - 1) % i != 0)
               return false;
           n = n / i;
        
    }
     
    // This condition is to handle
    // the case when n
    // is a prime number > 2
    if (n > 2)
        if ((N / n - 1) % n != 0)
            return false;
             
    return true;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Number N
    int n = 30;
 
    // Function Call
    if (isGiugaNum(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Pratima Pandey


Python3




# Python program for the above approach
import math
 
# Function to check if n is
# a composite number
def isComposite(n):
 
    # Corner cases
    if(n <= 1):
        return False
    if(n <= 3):
        return False
 
    # This is checked to skip
    # middle 5 numbers
    if(n % 2 == 0 or n % 3 == 0):
        return True
 
    i = 5
    while(i * i <= n):
        if(n % i == 0 or n % (i + 2) == 0):
            return True
        i += 6
    return False
 
# Function to check if N is a
# Giuga Number
def isGiugaNum(n):
 
    # N should be composite to be
    # a Giuga Number
    if(not(isComposite(n))):
        return False
    N = n
 
    # Print the number of 2s
    # that divide n
    while(n % 2 == 0):
        if((int(N / 2) - 1) % 2 != 0):
            return False
        n = int(n / 2)
 
    # N must be odd at this point.
    # So we can skip one element
    for i in range(3, int(math.sqrt(n)) + 1, 2):
         
        # While i divides n, 
           # print i and divide n
        while(n % i == 0):
            if((int(N / i) - 1) % i != 0):
                return False
            n = int(n / i)
 
    # This condition is to handle 
    # the case when n 
    # is a prime number > 2
    if(n > 2):
        if((int(N / n) - 1) % n != 0):
            return False
    return True
 
# Driver code 
 
# Given Number N
n = 30
if(isGiugaNum(n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by avanitrachhadiya2155


C#




// C# program for the above approach
using System;
class GFG{
     
// Function to check if n is
// a composite number
static bool isComposite(int n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return false;
     
    // This is checked to skip
    // middle 5 numbers
    if (n % 2 == 0 || n % 3 == 0)
        return true;
     
    for(int i = 5; i * i <= n; i = i + 6)
       if (n % i == 0 || n % (i + 2) == 0)
           return true;
     
    return false;
}
     
// Function to check if N is a
// Giuga Number
static bool isGiugaNum(int n)
{
     
    // N should be composite to be
    // a Giuga Number
    if (!(isComposite(n)))
        return false;
     
    int N = n;
     
    // Print the number of 2s
    // that divide n
    while (n % 2 == 0)
    {
        if ((N / 2 - 1) % 2 != 0)
            return false;
        n = n / 2;
    }
     
    // N must be odd at this point.
    // So we can skip one element
    for(int i = 3; i <= Math.Sqrt(n);
            i = i + 2)
    {
        
       // While i divides n,
       // print i and divide n
       while (n % i == 0)
       {
           if ((N / i - 1) % i != 0)
               return false;
           n = n / i;
       }
    }
     
    // This condition is to handle
    // the case when n
    // is a prime number > 2
    if (n > 2)
        if ((N / n - 1) % n != 0)
            return false;
     
    return true;
}
 
// Driver code
static void Main()
{
         
    // Given Number N
    int N = 30;
     
    // Function Call
    if (isGiugaNum(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by divyeshrabadiya07


Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if n is
// a composite number
function isComposite(n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return false;
 
    // This is checked to skip
    // middle 5 numbers
    if (n % 2 == 0 || n % 3 == 0)
        return true;
 
    for(let i = 5; i * i <= n; i = i + 6)
       if (n % i == 0 || n % (i + 2) == 0)
           return true;
 
    return false;
}
 
// Function to check if N is a
// Giuga Number
function isGiugaNum(n)
{
     
    // N should be composite to be
    // a Giuga Number
    if (!(isComposite(n)))
        return false;
 
    let N = n;
 
    // Print the number of 2s
    // that divide n
    while (n % 2 == 0)
    {
        if ((N / 2 - 1) % 2 != 0)
            return false;
             
        n = n / 2;
    }
 
    // N must be odd at this point.
    // So we can skip one element
    for(let i = 3;
            i <= Math.sqrt(n);
            i = i + 2)
    {
        
       // While i divides n,
       // print i and divide n
       while (n % i == 0)
       {
           if ((N / i - 1) % i != 0)
               return false;
                
           n = n / i;
        
    }
     
    // This condition is to handle
    // the case when n
    // is a prime number > 2
    if (n > 2)
        if ((N / n - 1) % n != 0)
            return false;
             
    return true;
}
 
// Driver Code
 
// Given Number N
let n = 30;
 
// Function Call
if (isGiugaNum(n))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by susmitakundugoaldanga     
 
</script>


Output: 

Yes

 



Last Updated : 07 Jul, 2021
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